1437. Check If All 1's Are at Least Length K Places Away (E)

题目

Given an array nums of 0s and 1s and an integer k, return True if all 1’s are at least k places away from each other, otherwise return False.

Example 1:

Input: nums = [1,0,0,0,1,0,0,1], k = 2
Output: true
Explanation: Each of the 1s are at least 2 places away from each other.

Example 2:

Input: nums = [1,0,0,1,0,1], k = 2
Output: false
Explanation: The second 1 and third 1 are only one apart from each other.

Example 3:

Input: nums = [1,1,1,1,1], k = 0
Output: true

Example 4:

Input: nums = [0,1,0,1], k = 1
Output: true

Constraints:

• 1 <= nums.length <= 10^5
• 0 <= k <= nums.length
• nums[i] is 0 or 1

题意

给定一个只包含0和1的数组，判断每个1之间的间隔是否大于指定值。

思路

遍历的同时判断即可。

代码实现

Java

class Solution {
    public boolean kLengthApart(int[] nums, int k) {
        int pre = -k - 1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 1) {
                if (i - pre <= k) return false;
                pre = i;
            }
        }
        return true;
    }
}